Question: Three of the vertices of parallelogram $ABCD$ are $A = (3,-1,2),$ $B = (1,2,-4),$ and $C = (-1,1,2).$  Find the coordinates of $D.$
Explanation: Since $ABCD$ is a parallelogram, the midpoints of diagonals $\overline{AC}$ and $\overline{BD}$ coincide.

[asy]
unitsize(0.4 cm);

pair A, B, C, D;

A = (0,0);
B = (7,2);
D = (1,3);
C = B + D;

draw(A--B--C--D--cycle);
draw(A--C,dashed);
draw(B--D,dashed);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);

dot((A + C)/2);
[/asy]

The midpoint of $\overline{AC}$ is
\[\left( \frac{3 + (-1)}{2}, \frac{(-1) + 1}{2}, \frac{2 + 2}{2} \right) = (1,0,2).\]This is also the midpoint of $\overline{BD},$ so the coordinates of $D$ are
\[(2 \cdot 1 - 1, 2 \cdot 0 - 2, 2 \cdot 2 - (-4)) = \boxed{(1,-2,8)}.\]